Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $r = \dfrac{k + 5}{6k + 18} \div \dfrac{k - 2}{k^2 + k - 6} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{k + 5}{6k + 18} \times \dfrac{k^2 + k - 6}{k - 2} $ First factor the quadratic. $r = \dfrac{k + 5}{6k + 18} \times \dfrac{(k + 3)(k - 2)}{k - 2} $ Then factor out any other terms. $r = \dfrac{k + 5}{6(k + 3)} \times \dfrac{(k + 3)(k - 2)}{k - 2} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ (k + 5) \times (k + 3)(k - 2) } { 6(k + 3) \times (k - 2) } $ $r = \dfrac{ (k + 5)(k + 3)(k - 2)}{ 6(k + 3)(k - 2)} $ Notice that $(k - 2)$ and $(k + 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ (k + 5)\cancel{(k + 3)}(k - 2)}{ 6\cancel{(k + 3)}(k - 2)} $ We are dividing by $k + 3$ , so $k + 3 \neq 0$ Therefore, $k \neq -3$ $r = \dfrac{ (k + 5)\cancel{(k + 3)}\cancel{(k - 2)}}{ 6\cancel{(k + 3)}\cancel{(k - 2)}} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $r = \dfrac{k + 5}{6} ; \space k \neq -3 ; \space k \neq 2 $